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The Big Division (Posted on 2019-11-18) Difficulty: 3 of 5
What is the remainder when 232323...232323 (4018 digits) is divided by 999?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution | Comment 1 of 4
991

The following program simulates long division taking the digits 2 at a time, but only keeps track of the remainder.

digits = 4018
location = 0
remainder = 0
while digits>0:
    divisor = remainder*100 + 23
    remainder = divisor % 999
    location += 2
    digits -= 2

print(location,remainder)

Output:  4018 991

----
Also note that the remainders follow a repeating pattern of period 27.
(but if you did this one digit at a time, I expect the cycle would be of period 54)

  Posted by Larry on 2019-11-18 08:11:17
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